There are 5 green 7 red balls. Two balls are selected one by one without replacement. Find the probability that first is green and second is red.

Let A = first ball is green, B = second ball is red.

We need to find to the probability that first ball is green and second ball is red:

$P(A\cap B)=P(A)\cdot P(B|A)$

There are 12 balls and 5 of them are green. So, $P(A)=\frac{5}{12}$ .

Since 1 green ball is selected, there are 11 balls and 7 of them are red. So, $P(B|A)=\frac{7}{11}$ .

$P(A\cap B)=\frac{5}{12}\cdot \frac{7}{11}=\frac{35}{132}$ .

Answer: $\frac{35}{132}$ .